Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

FLUID MECHANICS PLEASE HELP?

Water stands at a depth H in a large open tank whose side walls are vertical. A hole is made in one of the walls at a depth h below the water surface.

At what distance R from the foot of the wall does the emerging stream strike the floor? 2sqrt(h(H-h))

How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole?????

3 Answers

Relevance
  • 1 decade ago
    Favorite Answer

    At the hole the speed of the water is : v = sqr{2 g(H-h)} horizontal.

    R = v*t = v*sqr{2h/g} = sqr{2g(H-h)*2h/g} = 2 sqr{h(H-h)}

    h' = h

  • Anonymous
    1 decade ago

    for the question just use ur formulas!

    for the second question: depth H

  • Anonymous
    1 decade ago

    go to the internet and type it in to find out the answer

    Source(s): internet
Still have questions? Get your answers by asking now.